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LynFran
 one year ago
tan(1/2(cos^−1(x))=√1−x/1+x
LynFran
 one year ago
tan(1/2(cos^−1(x))=√1−x/1+x

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you have any idea on starting this question?

LynFran
 one year ago
Best ResponseYou've already chosen the best response.1no is say prove that their equal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well that's the first step :) Do you know what to do next?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We don't need to rush in and do everything in one go :) We can firstly start of with: tan(1/2(cos^−1(x))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Tan ( 1 ) 2 Cos^1(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Try changing the inverse trig ratio

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You get something like: Tan (Secx) 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So graphically, it looks something like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433678743148:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The only way to get the prove this, is by using properties of Pythagoras's, since we are dealing with a right angle triangle

LynFran
 one year ago
Best ResponseYou've already chosen the best response.1ok so we finding the hypotenuse in other word?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah we could try to find the hypotenuse or the other side. Just some way to get us that Square root, to help us prove that LHS = RHS

LynFran
 one year ago
Best ResponseYou've already chosen the best response.1so we would have x^2=(1/cosx)^2+2^2)?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Question: \(tan (\dfrac{1}{2} arccos (x))\) or \(tan (\dfrac{1}{2arccos (x))}\). which one?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0if so, then the right hand side must be \(\dfrac{\sqrt {1x^2}}{1+x}\) , right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0you missed the square at x from numerator. Am I right? please check the problem

LynFran
 one year ago
Best ResponseYou've already chosen the best response.1the right side has no square at x its just; the root of (1x)/(1+x)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0If it is so, I failed. :)
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